TITLE: Isometries of some simple Cayley graphs
QUESTION [12 upvotes]: Consider a Cayley graph of a group $G$ with respect to a symmetric finite generating set $S$.
There are some obvious candidates to isometries of this graph - for example, translation by elements of $G$, and group automorphisms which preserve $S$.
In some simple cases, these are everything one has (up to composition) - for example, take $G$ be a the free abelian group $G=\mathbb{Z}^{d}$ with $S$ being $\{\pm e_i\}$, with $\{e_i\}$ being the standard basis. This is true for other generating sets as well.
However, it is far from correct, say, for the free group with its standard generators.
I am more inclined toward "$\mathbb{Z}^d$-esque" situations in my research, so I would like to try and know more on the first case example.
For the questions themselves:
(a) Consider $G=\mathbb{Z}^d$ with some generating set. Can all isometries of the Cayley graph be seen as a composition of translations and group Automorphisms (preserving S)?
(b) Is this claim true for "nice" nilpotent groups, e.g. the Heisenberg groups?
(c) Must it be false for hyperbolic groups, considering all possible generating sets?

REPLY [10 votes]: Question (b) (and, therefore, also Question (a)) was answered affirmatively (for all torsion-free nilpotent groups) in 1998 by R.G.Möller and N.Seifter [Europe J. Comb. 19, 597-602] (see Theorem 4.1(1)).
The answer to Question (a) was rediscovered in 2007 by A.A.Ryabchenko [Siberian Math. J. 48 (2007) 919–922] (see the proof of Theorem 2). This short proof is reproduced in Theorem 5.3 of a preprint of J.Morris (arXiv:1502.06114).
A more elementary proof of the answer to Question (b) (for all torsion-free nilpotent groups) is in a preprint by J.Morris, G.Verret, and me (arXiv:1603.01883).
I have no information about Question (c).